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I am looking for a correct proof of this statement: If $G$ is a group such that $G/Z(G)$ is cyclic, then $G$ is commutative.

Proof: $G/Z(G)$ is isomorphic to $\operatorname{Inn}(G)$ and is cyclic, and then for every $a$ and $b$ in $G$ the inner isomorphisms $\gamma_a$ and $\gamma_b$ satisfy $\gamma_a \gamma_b = \gamma_{ab} = \gamma_{ba} = \gamma_b \gamma_a$, and therefore for every $a,b \in G$, $ab = ba$.

Is that proof complete, or am I missing something? Thanks a lot for the help.

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marked as duplicate by Jonas Meyer, BlackAdder, ᴡᴏʀᴅs, Daniel W. Farlow, Claude Leibovici Mar 31 '15 at 5:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1. There's something wrong with your therefore step: What you've written implies $aba^{-1}b^{-1}$ lies in the center only. 2. You only used Inn(G) is abelian, and this does not imply $G$ is commutative. – Soarer Nov 29 '10 at 9:36
You really don't need to touch automorphisms for this one. – Alex B. Nov 29 '10 at 9:42
up vote 3 down vote accepted

A problem with your attempt is that you seem to want to use only the fact that the inner automorphism group is abelian, but this does not suffice. There are nonabelian groups $G$ such that $G/Z(G)$ is abelian, like the group of symmetries of the square. Thus, it does not follow from $\gamma_{ab}=\gamma_{ba}$ that $ab=ba$.

So you need to use the hypothesis that $G/Z(G)$ has a single generator. Robin Chapman pointed out to you here what you can conclude from this, so I might as well quote him:

Each element of $G$ has the form $a^nz$ where $a$ is fixed and $z\in Z(G)$...

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Ok - we know, that there is element $a \in G$ for which is $<aZ(G)>=G/Z(G)$, and that give us path to express any element $g \in G$ in the form $a^n z$, great, that I now see... And therefore for $g, h \in G$ we give $gh = a^n z a^m z'$ and becouse $z$ is from $Z(G)$ we give $a^n z a^m z' = a^m z' a^n z = hg$. Great - thanks a lot. :-) – tomas.lang Nov 29 '10 at 10:08

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